제출 #1203029

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1203029		zeta7532	은하철도 (APIO24_train)	C++20	0 / 100	159 ms	23920 KiB

train

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
static const ll INF = (1LL<<60);

// Persistent Segment Tree for point-add & range-sum
struct PST {
    struct Node { int l, r, sum; };
    vector<Node> st;
    int N;
    PST(int _N): N(_N) {
        st.reserve(_N * 20);
        st.push_back({0,0,0});  // dummy at idx=0
    }
    // build empty tree on [l..r], return root index
    int build(int l, int r) {
        int id = st.size();
        st.push_back({0,0,0});
        if (l < r) {
            int m = (l + r) >> 1;
            st[id].l = build(l, m);
            st[id].r = build(m+1, r);
        }
        return id;
    }
    // create new version from prev, add v at pos p
    int update(int prev, int l, int r, int p, int v) {
        int id = st.size();
        st.push_back(st[prev]);
        if (l == r) {
            st[id].sum += v;
        } else {
            int m = (l + r) >> 1;
            if (p <= m)
                st[id].l = update(st[prev].l, l, m, p, v);
            else
                st[id].r = update(st[prev].r, m+1, r, p, v);
            st[id].sum = st[st[id].l].sum + st[st[id].r].sum;
        }
        return id;
    }
    // query sum on [ql..qr]
    int query(int id, int l, int r, int ql, int qr) const {
        if (!id || qr < l || r < ql) return 0;
        if (ql <= l && r <= qr) return st[id].sum;
        int m = (l + r) >> 1;
        return query(st[id].l, l, m, ql, qr)
             + query(st[id].r, m+1, r, ql, qr);
    }
};

// signature must match exactly by-value vectors
long long solve(int n, int m, int w,
                vector<int> t,
                vector<int> x,
                vector<int> y,
                vector<int> a,
                vector<int> b,
                vector<int> c,
                vector<int> L,
                vector<int> R) 
{
    // 1) 時刻を収集して圧縮
    vector<int> allT;
    allT.reserve(2*m + 2*w + 1);
    allT.push_back(0);
    for (int i = 0; i < m; i++) {
        allT.push_back(a[i]);
        allT.push_back(b[i]);
    }
    for (int i = 0; i < w; i++) {
        allT.push_back(L[i]);
        allT.push_back(R[i]);
    }
    sort(allT.begin(), allT.end());
    allT.erase(unique(allT.begin(), allT.end()), allT.end());
    auto comp = [&](int v){
        return int(lower_bound(allT.begin(), allT.end(), v) - allT.begin());
    };
    int Tsz = allT.size();

    // 2) meals を r_comp ごとにバケット化
    vector<vector<int>> meals_at_r(Tsz);
    for (int i = 0; i < w; i++) {
        int lc = comp(L[i]), rc = comp(R[i]);
        meals_at_r[rc].push_back(lc);
    }

    // 3) 永続セグ木のバージョンを構築
    PST pst(Tsz);
    vector<int> version(Tsz);
    version[0] = pst.build(0, Tsz-1);
    for (int lpos : meals_at_r[0]) {
        version[0] = pst.update(version[0], 0, Tsz-1, lpos, 1);
    }
    for (int i = 1; i < Tsz; i++) {
        version[i] = version[i-1];
        for (int lpos : meals_at_r[i]) {
            version[i] = pst.update(version[i], 0, Tsz-1, lpos, 1);
        }
    }

    // g(b,a): 区間 (b,a) に完全包含される meal の数
    auto get_g = [&](int bval, int aval){
        int bc = comp(bval), ac = comp(aval);
        if (ac == 0) return 0;
        int tot = pst.query(version[ac-1], 0, Tsz-1, 0, Tsz-1);
        int bad = pst.query(version[ac-1], 0, Tsz-1, 0, bc);
        return tot - bad;
    };

    // 4) 列車を出発時刻 a の昇順で処理する順序
    vector<int> idx(m);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(),
         [&](int i, int j){ return a[i] < a[j]; });

    // DP 配列
    vector<ll> dp(m, INF);

    // 各惑星ごとに到着候補を heap & deque で管理
    vector< priority_queue<pair<int,int>,
             vector<pair<int,int>>, greater<pair<int,int>>> > heap(n);
    vector< deque<int> > dq(n);

    // 初期状態：planet 0, time 0 => j = -1
    heap[0].push({0, -1});
    auto get_dp = [&](int j){ return j < 0 ? 0LL : dp[j]; };
    auto get_b  = [&](int j){ return j < 0 ? 0    : b[j]; };

    ll ans = INF;
    // 5) departure 昇順で DP
    for (int _k = 0; _k < m; _k++) {
        int i = idx[_k];
        int p = x[i], ai = a[i];
        auto &hp = heap[p];
        auto &dq_p = dq[p];

        // step1: 到着済み j を dq に移動
        while (!hp.empty() && hp.top().first <= ai) {
            int j = hp.top().second;
            hp.pop();
            // pop-back by quadrangle-inequality
            while (dq_p.size() >= 2) {
                int j2 = dq_p.back(), j1 = dq_p[dq_p.size()-2];
                ll lhs = get_dp(j2) + (ll)get_g(get_b(j2), ai) * t[p];
                ll rhs = get_dp(j1) + (ll)get_g(get_b(j1), ai) * t[p];
                if (lhs >= rhs) dq_p.pop_back();
                else break;
            }
            dq_p.push_back(j);
        }

        // step2: dq_p.front() が最適遷移元
        if (!dq_p.empty()) {
            int j = dq_p.front();
            dp[i] = get_dp(j)
                  + (ll)get_g(get_b(j), ai) * t[p]
                  + c[i];
            if (y[i] == n-1) ans = min(ans, dp[i]);
        }

        // step3: arrival heap に追加
        heap[y[i]].push({b[i], i});
    }

    if (ans >= INF/2) return -1;
    return ans;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 30
• Subtask 4: 0 / 60